What is the second derivative of 1/(1+x^2)?

Dec 12, 2016

${d}^{\left(2\right)} / \left({\mathrm{dx}}^{2}\right) \left(\frac{1}{1 + {x}^{2}}\right) = \frac{6 {x}^{2} - 2}{{\left(1 + {x}^{2}\right)}^{3}}$

Explanation:

$f \left(x\right) = \frac{1}{1 + {x}^{2}}$

Based on the chain rule:

$f ' \left(x\right) = \frac{d}{\mathrm{dx}} {\left(1 + {x}^{2}\right)}^{- 1} = \left(2 x\right) \left(- 1\right) {\left(1 + {x}^{2}\right)}^{- 2} = \frac{- 2 x}{{\left(1 + {x}^{2}\right)}^{2}}$

Using the quotient rule:

$f ' ' \left(x\right) = \frac{\left(- 2\right) {\left(1 + {x}^{2}\right)}^{2} + 2 x \cdot 2 x \cdot 2 \left(1 + {x}^{2}\right)}{{\left(1 + {x}^{2}\right)}^{4}} = \frac{\left(- 2\right) \left(1 + {x}^{2}\right) + 8 {x}^{2}}{{\left(1 + {x}^{2}\right)}^{3}} = \frac{6 {x}^{2} - 2}{{\left(1 + {x}^{2}\right)}^{3}}$